Which among the following is a formula for force when the jet strikes the curved plate at the Centre to the direction of flow?

Q1) Find the force exerted by a jet of water of diameter 75 mm on a stationary flat plate, when the jet strikes the plate normally with a velocity of 20 m/s.

Solution: Given:

i) Diameter of jet, d = 75mm = 0.075mm

ii) Area, $a = \dfrac\pi4d^2 = \dfrac\pi4(0.075)^2 = 0.004417m^2$

iiii) Velocity of jet, v = 20m/s

The force exerted by the jet of water on a stationary plate is given by,

$F = \rho av^2$

$F = 1000\times 0.004417\times20^2 = 1766.8N$

Q2) A jet of water of diameter 50 mm strikes a fixed blade in such a way that the angle between the plate and the jet is $30^\circ$. The force exerted in the direction of jet is 1471.5N. Determine the rate of flow of water.

Solution:

Given:

i) Diameter of jet, d = 50mm = 0.05m

ii) Area, $a = \dfrac\pi4\times(0.05)^2 = 0.00163m^2$

iii) Angle,$\theta = 30^\circ$

iv) Force in the direction of jet,

$F_x = 1471.5N$

As force is given in (N), $\rho$ should be taken equal 1000 kg/m$^3$

$\therefore 1471.5 = 1000\times0.001963\times v^2\times \sin^230$

$\therefore v^2 = \dfrac {150}{0.05} = 3000$

$v = \sqrt{3000} = 54.77m/s$

Discharge, Q = Area $\times$ velocity

$Q=0.001963\times 54.77=0.1075m^3/s$

Q3) A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet of water in the direction of the jet, if the jet is deflected through an angle of 120$^\circ$ at the outlet of the curved plate.

Solution:

Given:

i) Diameter of jet, d = 50mm = 0.05m

ii) Area, $a = \dfrac \pi4(0.05)^2 = 0.001963m^2$

iii) Velocity, v = 40m/s

$\therefore$ To find $\theta$, use equation

$180-\theta$ = given angle

$180-\theta = 120$

$\therefore, \theta = 180-120 = 60^\circ$

$\therefore F_x = \rho av^2[1+\cos\theta]$

$F_x = 1000\times 0.001963\times40^2\times[1+\cos 60^\circ]$

$F_x = 4711.5N$

Q4) A jet of water of diameter 75 mm moving with a velocity of 30 m/s, strikes a curved fixed plate tangentially at one end at an angle of 30$^\circ$ to the horizontal, the jet leaves the plate at an angle of 20$^\circ$ to the horizontal, find the force exerted by a jet on the plate in the horizontal and vertical direction.

Solution:Given:

i) Diameter of jet, d = 75mm = 0.075m

ii) Area, $a = \dfrac\pi4\times(d^2)$

$a = \dfrac\pi4\times(0.075)^2 = 0.004417m^2$

iii) Velocity, v = 30m/s

iv) Angle made by the jet at inlet tip with horizontal, $\theta = 30^\circ$

v) Angle made by the jet at outlet tip with horizontal, $\phi = 20^\circ$

$\therefore F_x = \rho av^2[\cos\theta+\cos\phi]$

$F_x = 1000\times0.004417\times30^2\times[\cos30^\circ+\cos20^\circ]$

$F_x = 7178.2N$

$F_y = \rho av^2[\sin\theta-\sin\phi]$

$F_y = 1000\times0.004417[\sin30-\sin20]\times30^2$

$F_y = 628.13N$

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Option 1 : ρa(v - u)2 × u [1 + cos θ]

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20 Questions 20 Marks 20 Mins

Explanation:

Force exerted by a jet on a curved plate when the plate is moving in the direction of jet
Let us consider a jet of water striking a curved plate at the center of the plate which is moving with a uniform velocity in the direction of jet as displayed here in following figure.

Let us consider the following terms from above figure
V = Absolute velocity of jet, a = Area of jet, u = Velocity of the plate in the direction of jet
The relative velocity of water jet or velocity with which the water jet strikes the curved plate = (V - u)
Let us consider that plate is smooth and the loss of energy due to the impact of jet is zero. We will have a similar velocity with which the jet will be leaving the curved plate i.e. (V -u)
Velocity will be resolved into two components. One component will be in the direction of the jet and the second component will be in the direction perpendicular to the jet.
Component of the velocity in the direction of jet = - (V - u) Cos θ
We have taken a negative sign as at the outlet, the component will be in the opposite direction of jet.
Component of the velocity in the direction perpendicular to the jet = (V - u) Sin θ
Mass of the water striking the plate = ρ a (V - u)

Force exerted by the jet of water on the curved plate in the direction of jet

Fx = Mass striking per second × [Initial velocity with which jet strikes the plate in the direction of jet - Final velocity]

Fx = ρa(V - u) [(V - u) - (- (V - u) cos θ)]

Fx = ρa(V - u) [(V - u) + (V - u) cos θ]

Fx = ρa(V - u)2 [1 + cos θ]

Work done by the jet per second on the plate will be given by following equation

W = Fx × Distance travelled per second in the direction of x

W = Fx × u = a(V - u)2 [1 + cos θ] × u

W = ρa(V - u)2 × u [1 + cos θ]

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